High School

How much energy is required to vaporize 158 g of butane (C₄H₁₀) at its boiling point, if its ΔHvap is 24.3 kJ/mol?

A. 15.1 kJ
B. 66.1 kJ
C. 89.4 kJ
D. 11.2 kJ
E. 38.4 kJ

Answer :

Final answer:

To find the energy required to vaporize butane, we use the equation: Energy (kJ) = mole of butane * ΔHvap. Plugging in the values, we can determine that the energy required is 66.1 kJ. The correct answer is B) 66.1 kJ.

Explanation:

To calculate the energy required to vaporize butane, we need to use the equation:

Energy (kJ) = mole of butane * ΔHvap

First, we need to find the number of moles of butane:
Number of moles = mass of butane / molar mass of butane

Molar mass of butane (C4H10) = 4(12.01 g/mol) + 10(1.01 g/mol) = 58.12 g/mol

Number of moles of butane = 158 g / 58.12 g/mol = 2.72 mol

Now, we can calculate the energy required:

Energy (kJ) = 2.72 mol * 24.3 kJ/mol = 66.1 kJ

Therefore, the correct answer is B) 66.1 kJ.

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